## 17Calculus Limits - Solve By Substitution

##### 17Calculus

Evaluating Limits Using The Technique of Substitution

### Finite Limits, Infinite Limits, Limits At Infinity . . . Terminology Explained

The use of the terms finite limits, infinite limits and limits at infinity are used differently in various books and your instructor may have their own idea of what they mean. In this panel, we will try to break down the cases and explain the various ways these terms can be used as well as how we use them here at 17calculus.

When we talk about limits, we are looking at the $$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$. The various terms apply to the description of $$c$$ and $$L$$ and are shown in the table below. The confusion lies with the terms finite limits and infinite limits. They can mean two different things.

$$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$

when

term(s) used

$$c$$ is finite

limits approaching a finite value or finite limits

$$c$$ is infinite $$\pm \infty$$

limits at infinity or infinite limits

$$L$$ is finite

finite limits

$$L$$ is infinite $$\pm \infty$$

infinite limits

You can see where the confusion lies. The terms finite limits and infinite limits are used to mean two different things, referring to either $$c$$ or $$L$$. It is possible to have $$c = \infty$$ and $$L$$ be finite. So is this an infinite limit or a finite limit? It depends if you are talking about $$c$$ or $$L$$.

How 17calculus Uses These Terms
The pages on this site are constructed based on what $$c$$ is, i.e. we use the terms finite limits and infinite limits based on the value of $$c$$ only ( using the first two rows of the table above and ignoring the last two ). This seems to be the best way since, when we are given a problem, we can't tell what $$L$$ is until we finish the problem, and therefore we are unable to determine what type of problem we have and know what techniques to use until we are done with the problem.

Important: Make sure and check with your instructor to see how they use these terms.

This is the first and most important technique for evaluating limits. It is very simple but you need to understand when you can use it.

Technique 1 - Substitution

We know from the concept of continuity, that if a function, $$f(x)$$, is continuous at the point $$x=c$$, then $$\displaystyle{ \lim_{x \to c}{[f(x)]} = f(c) }$$. So this tells us, that if the function is continuous at the point at which we are taking the limit, then we can just substitute that point into the function to determine the limit and we are done.

We can also substitute the limit value into the function to help us determine what to do next, even if the function is not continuous at the point. Here is what substitution tells us.

Case 1: After substitution, if we get an indeterminate form, we need to use one (or more) of the techniques discussed on other pages (factoring, rationalizing, using trig identities or, if you know derivatives, L'Hôpital's Rule).

Case 2: After substitution, if we get a fraction like $$c/0$$ where $$c$$ is any non-zero finite number this indicates a possible asymptote, so we need to look at what is happening on either side of $$x=c$$. There are 4 possibilities. (These graphs look at $$c=0$$. However, the discussion holds regardless of the value of $$c$$.)

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$

The limit exists and $$\displaystyle{ \lim_{x \to c}{[f(x)]} = +\infty }$$.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$

The limit exists and $$\displaystyle{ \lim_{x \to c}{[f(x)]} = -\infty }$$.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$

The limit does not exist.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$

The limit does not exist.

Note - On this site (and in many textbooks), we say that when $$\displaystyle{ \lim_{x \to c}{[f(x)]} = \infty }$$, the limit exists and the limit is infinity (similarly for negative infinity). Some textbooks and instructors say that the limit does not exist. Check with your instructor to see what they require.

Now let's use this technique on some practice problems.

Practice

Use the technique of substitution to solve these limits, giving your answers in exact, simplified form.

$$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}$$

$$\displaystyle{\lim_{x\rightarrow 3}\left[\frac{5x^2-8x-13}{x^2-5}\right]=2}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}$$

Solution

 You always want to try direct substitution first. $$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{5x^2-8x-13}{x^2-5} \right] }$$ $$\displaystyle{\frac{5(9)-8(3)-13}{9-5} }$$ $$\displaystyle{\frac{45-24-13}{4} }$$ $$\displaystyle{\frac{8}{4} = 2 }$$ In this case, the result is determinate. So the limit is 2.

$$\displaystyle{\lim_{x\rightarrow 3}\left[\frac{5x^2-8x-13}{x^2-5}\right]=2}$$

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$$\displaystyle{\lim_{x\to1}{\frac{3x+5}{x+4}}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\to1}{\frac{3x+5}{x+4}}}$$

Solution

### PatrickJMT - 22 video solution

video by PatrickJMT

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$$\displaystyle{\lim_{x\to0}{(x^2+x-6)}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\to0}{(x^2+x-6)}}$$

Solution

### PatrickJMT - 23 video solution

video by PatrickJMT

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$$\displaystyle{\lim_{x\to14}{(x+\sqrt{x+11})}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\to14}{(x+\sqrt{x+11})}}$$

Solution

### PatrickJMT - 24 video solution

video by PatrickJMT

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$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }$$

$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} = 2 }$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }$$

Solution

### Krista King Math - 1921 video solution

video by Krista King Math

$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} = 2 }$$

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$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}$$

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}=+\infty}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}$$

Solution

Direct substitution yields $$\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]} = \frac{-3}{0}}$$. So the answer is $$+\infty$$, $$-\infty$$ or DNE (Does Not Exist). How do you know which one (without a graph)?
We need to determine the behavior of the function on either side of $$x=0$$. On the left side of zero when $$x < 0$$ and is very close to zero, the numerator is negative. We know this since x is very, very small, so $$x^4$$ is positive but very small and $$5x$$ is very small but negative. Both terms will be dominated by -3, so the numerator is negative. In the denominator, we need to look at the square root; $$x^2$$ is always positive and adding a positive number to 4 means that the number under the square root is always greater than 4, so the square root is always larger than 2. This makes the denominator negative. So we have a negative numerator and a negative denominator, so the limit $$\displaystyle{ \lim_{x \rightarrow 0^-}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }$$
We use similar logic when $$x > 0$$, to get $$\displaystyle{ \lim_{x \rightarrow 0^+}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }$$

Since the limit from the right is equal to the limit from the left, the original limit is also $$+\infty$$.

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}=+\infty}$$

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When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

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